Statistics / Midterm Ex Exam (D (Duration: 90 90 mi minutes)
Name:
Q1. A long-distance taxi service owns four vehicles. These are of different ages and have
different repair records. The probabilities that on any given day, each vehicle will be available for use are .95, .90, .90, and .80. Whether one vehicle is available is independent indep endent of whether any other vehicle is available. (a) (a) Find ind the the prob probab abil ilit ity y func functi tion on for for the the num numbe berr of veh vehic icle less avai availlable able for for use use on on a given day. (b) (b) Find Find the the exp expec ecte ted d num numbe berr of of veh vehic icle less ava avail ilab able le for for use use on a giv given en day. day. Find the standard deviation of the number of vehicles available for use on a given day. 0.0001, P(1)=0.00 004 41, P(2)=0.0 0.0571, P(3)=0.32 323 31, P(4)=0.61 615 56 b) 3.55 Answer: a) P(0)=0.0 c) 0.6225 Solution: a) P(0) = (1-0.95)(1-0.9)(1-0.9)(1-0.8) = 0.0001 P(1) = 0.95(1-0.9)(1-0.9)(1-0.8) + (1-0.95)0.9(1-0.9)(1-0.8) + (1-0.95)(1-0.9)0.9(1-0.8) + + (1-0.95)(1-0.9)(1-0.9)0.8 = 0.0041 P(2) = (0.95)(0.9)(1-0.9)(1-0.8) + (0.95)(1-0.9)(0.9)(1-0.8) + (0.95)(1-0.9)(1-0.9)(0.8) + (1-0.95) (0.9)(0.9)(1-0.8) + (1-0.95)(0.9)(1-0.9)(0.8) + (1-0.95)(1-0.9)(0.9)(0.8) = 0.0571 P(3) = (0.95)(0.9)(0.9)(1-0.8) + (0.95)(0.9)(1-0.9)(0.8) + (0.95)(1-0.9)(0.9)(0.8) + (1-0.95)(0.9) (0.9)(0.8) = 0.3231 P(4) = (0.95)(0.9)(0.9)(0.8) = 0.6156 b) E(X) = SUM(x.p(x)) = 3.55 STD(X) = SUM ((x-E(X)) 2 p(x)) = 0.6225 Q2. A team of five analysts is about to examine the earnings prospects of twenty corporations.
Each of the five analysts will study four of the corporations. These analysts are not equally competent. In fact, one of them is a star, having an excellent record of anticipating changing trends. Ideally, management would like to allocate to this analyst the four corporations whose earnings will deviate most from past trends. However, lacking this information, management allocates corporations to analysts randomly. What is the probability that at least two of the four corporations whose earnings will deviate most from past trends are allocated to the star analyst? Answer: 0.1808 Solution:
Let X be the number of the four corporations corporations whose earnings earnings will deviate most from past trends are allocated to the star analyst. Then X has a hypergeometric distribution: n=5 (number of trials), N=20 (population size) and s=4 (number of possible occurences). Need to find P(X≥2) = 1- P(0)- P(1) = 1 – 0.37564 – 0.46233 = 0.16202 Q3. It is estimated that amounts of money spent on gasoline by customers at a gas station follow
a normal distribution with standard deviation $2.50. It was also found that 10% of all customers spend more than $15. What percentage of customers spend less than $10? Answer: 23.58% Solution: X=N(µ,2.52) We know P(X>15) = 0.1, and want to find P(X<10). P(Z>(15-µ)/2.5) = 0.1 P(Z≤(15-µ)/2.5) = 0.9 Hence (15-µ)/2.5 = 1.28, so µ = 11.8. Thus, X=N(11.8,2.52). P(X<10) = 0.2358. Thus, 23.58% of customers spend less than $10. particular contract. contract. The contract contract Q4. A consultant knows that it will cost her $10,000 to fulfill a particular is to be put out for bids, and she believes that the lowest bid, excluding her own, can be represented represented by a distribution distribution that is uniform between $8,000 and $20,000. $20,000. Therefore, Therefore, if the ran-
dom variable X denotes the lowest of all other bids (in thousands of dollars), its probability density function is 1 / 12, for 8 < x < 20 f ( x) = for all other values of x 0, (a) What is the probability that the lowest of the other bids will be less than the consultant's cost estimate of $10,000? (b) If the consultant submits a bid of $12,000, what is the probability that she will secure the contract? (c) The consultant decides to submit a bid of $12,000. What is her expected profit from this strategy? (d) If the consultant wants to submit a bid so that her expected profit is as high as possible, discuss how she should go about making this choice. b) 2/3 c) $1,333.33 d) $15,000 Answer: a) 1/6 Solution: X = U(8,20) a) P(X<10) = (10-8)/(20-8) = 2/12 = 1/6 b) P(X>12) = (20-12)/12=8/12=2/3 c) Expected profit = (12-10)(20-12)/12 = 4/3 = 1,333.33 (thousand US dollars) d) Let B denote the bid (in thousand of dollars). Then the expected profit is (B-10)(20-B)/12 thousand dollars. Choose B to maximize this, giving B=$15,000. Q5. In a large department store, a customer complaints office handles an average of six
complaints per hour about quality of service. The distribution is Poisson. (a) What is the probability that in any hour exactly six complaints will be received? (b) What is the probability that more than 20 minutes will elapse between successive complaints? (c) What is the probability that less than 5 minutes will elapse between successive complaints? (d) The store manager observes the complaints office for a 30-minute period, during which no complaints are received. She concludes that a talk she gave to her staff on the theme "the customer is always right" has obviously had a beneficial effect. Suppose that in fact the talk had no effect. What is the probability of the manager's observing a period of 30 minutes or longer with no complaints? b) 0.135 c) 0.393 d) 0.05 Answer: a) 0.16 Solution: X=Poisson(6) a) Probability of exactly 6 complaints will be received = f(6) = 6 6 . e-6 / 6! = 0.16 The length of the interval between two occurences Y follows exponential distribution, with mean µ=10 (minutes). b) P(Y>20) = e -2 = 0.135 c) P(Y<5) = 1- e -1/2 = 0.393
